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Unformatted text preview: 1 C:\User\Teaching\505-506\S11\HW03'sol.doc PHY 506 Classical Electrodynamics II Spring 2011 Homework 3 with solutions Problem 3.1 (to be graded of 20 points). Use the recipe outlined in Sec. 7.8 of the lecture notes to derive the characteristic equation (7.157), 2 2 2 2 2 2 2 2 2 2 1 1 1 1 t t t t n ' n t n ' n t n ' n t n ' n t k k k k R n K K J J k K K k J J k k , for the HE and EH modes in a round, step-index optical fiber . Solution : Let us start with calculating the constant coefficient in Eq. (7.156) from the requirement that at the interface ( = R ), both E z and H z are continuous (as tangential field components), i.e. f + = f- . From here and Eq. (152) we get in e K R K R k J f f t n t n t n l ) ( ) ( ) ( . Now we can use the first of Eqs. (117), together with the general vector-algebra expressions for gradient’s components in polar coordinates, 1 which give , , 1 f f f f t z t z n n to obtain the following expressions for the transversal components of the electric field: , ) ( ) ( , ) ( ) ( 2 2 in in e H k J k Z k E k J in k k i E e H k J in Z k E k J k k k i E l t ' n t l t n z t l t n l t ' n t z t . ) ( ) (...
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